Question: The lifespans of tigers in a particular zoo are normally distributed. The average tiger lives $26.2$ years; the standard deviation is $3.2$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a tiger living less than $29.4$ years.
Answer: $26.2$ $23$ $29.4$ $19.8$ $32.6$ $16.6$ $35.8$ $68\%$ $16\%$ $16\%$ We know the lifespans are normally distributed with an average lifespan of $26.2$ years. We know the standard deviation is $3.2$ years, so one standard deviation below the mean is $23$ years and one standard deviation above the mean is $29.4$ years. Two standard deviations below the mean is $19.8$ years and two standard deviations above the mean is $32.6$ years. Three standard deviations below the mean is $16.6$ years and three standard deviations above the mean is $35.8$ years. We are interested in the probability of a tiger living less than $29.4$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $68\%$ of the tigers will have lifespans within 1 standard deviation of the average lifespan. The remaining $32\%$ of the tigers will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({16\%})$ will live less than $23$ years and the other half $({16\%})$ will live longer than $29.4$ years. The probability of a particular tiger living less than $29.4$ years is ${68\%} + {16\%}$, or $84\%$.